Quiz 8 Solution

Quiz 8 Solution

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Unformatted text preview: = 1 u6 + C = 1 (x2 + 4)6 + C. 6 6 ￿ √ 1 dx. 2x + 1 Solution. Let u = 2x + 1, du = 2 dx, 1 2 du = dx. Then ￿ 1 √ dx = 2x + 1 or equivalently, ￿ 1 √ · 1 du = u2 ￿ Exercise 3. Evaluate ￿3 5/2 2(2x √ 1 2 ￿ u−1/2 du = u1/2 + C = (2x + 1)1/2 + C, √ 1 dx = 2x + 1 + C. 2x + 1 − 5)14 dx. Solution. Let u = 2x − 5, If x = 5 , then u = 2( 5 ) − 5 = 0. 2 2 If x = 3, then u = 2(3) − 5 = 1. Then ￿3 5 /2 14 2(2x − 5) dx = ￿1 0 du = 2 dx. u14 du = ￿ 1 15 ￿1 u￿ 15 0 = 1 15 (1 − 0) = 1 15 ....
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This document was uploaded on 03/30/2014 for the course MATH 1142 at Minnesota.

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